JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 9)
A particle is subjected to two simple harmonic motions as :
$$
x_1=\sqrt{7} \sin 5 \mathrm{tcm}
$$
and $x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \mathrm{cm}$
where $x$ is displacement and $t$ is time in seconds.
The maximum acceleration of the particle is $x \times 10^{-2} \mathrm{~ms}^{-2}$. The value of $x$ is :
$5 \sqrt{7}$
125
$25 \sqrt{7}$
175
Explanation
$$\begin{aligned} &\begin{aligned} & x_1=\sqrt{7} \sin 5 t \\ & x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \end{aligned}\\ &\text { From phasor, } \end{aligned}$$
_2nd_April_Morning_Shift_en_9_1.png)
$$\begin{aligned} &\therefore \text { Amplitude of resultant } \mathrm{SHM}=7\\ &\begin{aligned} & \phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\ & \therefore X_R=7 \sin (5 \mathrm{t}+\phi) \\ & \quad a_R=-7 \times 25 \sin (5 \mathrm{t}+\phi) \\ & \therefore a_{\max }=175 \mathrm{~cm} / \mathrm{sec}=175 \times 10^{-2} \mathrm{~m} / \mathrm{sec} \end{aligned} \end{aligned}$$
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